The common tangent to the parabola \(y^2=32 x\) and \(x^2=256 y\) will be
The common tangent to the parabola \(y^2=32 x\) and \(x^2=256 y\) will be
\(2 x+4 y+64=0\)
\(x+2 y-32=0\)
\(2 x+4 y+32=0\)
\(4 x+2 y+64=0\)
Solution
Given parabolas are,
\(y^2=32 x \text { and } x^2=256 y\)
We use a standard result to find equation of common tangent. Equation of tangent common to \(y^2=4 a x\) and \(x^2=4 b y\) is,
\(b^{1 / 3} y+a^{1 / 3} x+\left(a^2 b^2\right)^{1 / 3}=0\)
Here,
\(\begin{aligned}
& 4 a=32 \Rightarrow a=8 \\
& 4 b=256, \quad b=64
\end{aligned}\)
So tangent is,
\(\begin{array}{ll}
& (64)^{1 / 3} y+(8)^{1 / 3} x+\left(8^2 \times 64^2\right)^{1 / 3}=0 \\
\Rightarrow \quad & 4 y+2 x+64=0
\end{array}\)